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11) The probability of an Emerald Gates subdivision resident having a privacy fence is 60%. The probability of a resident own
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Solution:

11) Let us define some events as follows:

A : A resident has a privacy fence

B : A resident has a dog

a) We have to obtain P(A|B).

Conditional probability of A given B is given as follows:

P(AB) = P(ANB) P(B)

P(A ∩ B) = Probability that a resident own a dog and also he has privacy fence

P(B) = Probability that a resident own a dog

We have, P(A ∩ B) = 25% = 0.25 and P(B) = 35% = 0.35

Hence, P(A|B) = 0.25/0.35 = 5/7

Given that a resident own a dog, the probability that he has a privacy fence is 5/7.

b) If two events A and B are independent then, P(A ∩ B) = P(A).P(B)

We have, P(A) = 0.60, P(B) = 0.35 and P(A ∩ B) = 0.25

Now, P(A).P(B) = 0.60×0.35 = 0.21 which is not equal to P(A ∩ B).

Hence, having a dog and having a fence are not independent events.

c) If two events A and B are disjoint events then P(A ∩ B) = 0

We are given that, P(A ∩ B) = 0.25 which is not equal to zero.

Therefore, having a dog and having a privacy fence are not disjoint events.

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